Lesson

Using
Pascal's
Triangle

Problem 1

Problem 1
Solution

Problem 2

Problem 2
Solution

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Pascal's Triangle

When numbers appear in arrays it is often possible to identify a pattern that enables us to continue the array indefintely.

Consider the following array of numbers which is known as Pascal's Triangle. Pascal's triangle has many applications in mathematics.

You can use patterns in the array to find the next few rows.

1. The first and last numbers in each row are always "1's"
   
   
2. The diagonal formed by the second entry in each row is just the counting numbers.
   
   
3. The diagonal formed by the third entry in each row begins 1 - 3 - 6 - 10 - . . .. Note that the difference between 1 and 3 is 2; the difference between 3 and 6 is 3; the difference between 6 and 10 is 4. We can conclude that the next number will be 5 more than 10 which is 15. And the next one after that will be 21 which is 6 more than 15.
   
   
4. The diagonal formed by the fourth entry in each row begins 1 - 4 - 10 - . . .. Note that the difference between 1 and 4 is 3; the difference between 4 and 10 is 6; These differences are in fact the numbers that appear in the diagonal of third entries! We can conclude that the next number will be 10 more than 10 which is 20. And the next one after that will be 35 which is 15 more than 20.
   
   
5. There is a symmetry in the array. Whatever number appears on the left appears in a corresponding position on the right. Thus we can complete a line of the array by finding the first half of the numbers in the array and then recopying thgem in reverse order. On alternate rows there is a middle number that is right in the middle and does not get repeated.

   Here are the next several rows of the array:

This array can be continued indefintely.

Besides the patterns that we identified in order to generate the next two rows, there are numerous other patterns that can be found in Pascal's Triangle. There is a particularly simple pattern that can be used to generate a row from the row before. We will use this pattern to generate the next row:

Begin with a 1.

To get the second entry, add the first two entries in the preceding row. [1 + 7 = 8]

To get the third entry, add the second and third entries in the preceding row. [7 + 21 = 28]

The fourth entry is gotten by adding the third and fourth entries in the preceding row. [21 + 35 = 56]

Continuing this process, and ending with a 1, we get the entries in the next row:

This process can be continued indefinitely in order to get as many rows as are needed.

See an example where Pascal's Triangle is used to solve a problem.

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email the author: Bruce Jacobs
Last modified: Thursday, February 6, 2003